Hey all having trouble with these questions:



The definition of integrability we use is the Riemann integrable definition, so if f is integrable on [a,b] then for all partitions of [a,b], the supremum of the lower sums is equal to the infimum of the upper sums.

For part (a) would I split the integral into an integral from -inf to 0 and 0 to +inf and then somehow show that the function is integrable on those intervals using the definition of Riemann integrability? Or is there an easier way?

Part (b) I can't really distinguish between the two integrals, to me they are the same.

And part (c) I'm not sure how to start. I'd like to make the argument that as N approaches inf then N+1 and N^2 don't really change the limit, but I suppose that is not rigorous enough.

b) The first integral does not exist as both the improper integral with zero diverge to infinity and -infinity, respectively. This means that the integral from -inf to +inf would be defined by
infinity - infinity
and this does not make sense; there is no consistent way of defining this.
On the other hand, the right integral (from -N to N) is 0 for every N and so converges to 0.

a) Instead of giving a real technical argument using partitions I would rather argue that, in case of the integral from 0 to +inf, the sequence of integrals from 0 to N converges, as it is bounded and strictly monotonic ([url]https://en.wikipedia.org/wiki/Monotone_convergence_theorem[/url])
- monotony: it's easy as the function itself is positive, and so is its integral on any domain
- bounded: you might consider the function 1/x^2 instead (however, the integral can't begin at 0, but at 1 for example. It doesn't destroy the argument). This has the advantage of having an easy antiderivative and as it is larger than the original function, its integral provides an upper bound for the integral of the other function.

c) If the integral from -inf to +inf exists this means that both the integrals [0, +inf) and (-inf, 0] exist, i.e. the sequences {integral on [0, N]} and {integral on [-N, 0]} both converge to the respective integrals. Then (using fantasy notation, I hope it's clear)
lim integral(-N, N) = lim (integral(-N, 0) + integral(0, N)) =(!)= lim integral(-N, 0) + lim integral(0, N) = integral(-inf, 0) + integral(0, inf) = integral(-inf, +inf)
For the step =(!)= you need the fact that both limits exist.

Concerning the second part of the question, I will have to think about this one.

For the second part of question c, you can replace the upper and lower bounds with functions u(N) and l(N) in my argument for the first part. Then, given that they are both monotonic and approach infinity and -inf, respectively (else they don't make much sense as bounds) the same argument as in the first part should really still hold.
Informally speaking, the reason is that as both the -inf to 0 and the 0 to +-inf part exist and converge it should in the limit make no difference which one is "faster" when considering their sum, as long as they both approach +/- infinity.
Apply all of this to u(N)=N+1 and l(N)=-N^2 to get the result for the given special cases.

This is very thorough, thanks a lot!

This is not necessary. All that matters is that both tends to infinity (u to + infinity and l to - infinity).

Yes you're right.